Results 1 to 3 of 3

Math Help - Find the Limit

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    32

    Find the Limit

    Hey all, sorry for bugging you again but I have another question...

    I've been doing them right so far, so I'm confused as to why this one's supposedly wrong.

    The question is: Evaluate the limit for:

    lim x-->3 squareroot(x+1)-2 / x-3

    So, I multipled the top and bottom by the conjugate of the numerator and got

    lim x-->3 (squareroot(x+1)-2)(squareroot(x+1)+2) / (x-3)(squareroot(x+1)+2)
    lim x-->3 (x+1)-4 / (x-3)(squareroot(x+1)+2)
    lim x-->3 (x-3) / (x-3)(squareroot(x+1)+2) and I cancelled the (x-3) from top and bottom

    I then plugged 3 into (squareroot(x+1)+2) to get (squareroot(4)+2) which is 2+2 = 4

    However, it says the answer is wrong. I'm really not sure why.

    Help is appreciated! Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member DeMath's Avatar
    Joined
    Nov 2008
    From
    Moscow
    Posts
    473
    Thanks
    5
    Quote Originally Posted by dark-ryder341 View Post
    Hey all, sorry for bugging you again but I have another question...

    I've been doing them right so far, so I'm confused as to why this one's supposedly wrong.

    The question is: Evaluate the limit for:

    lim x-->3 squareroot(x+1)-2 / x-3

    So, I multipled the top and bottom by the conjugate of the numerator and got

    lim x-->3 (squareroot(x+1)-2)(squareroot(x+1)+2) / (x-3)(squareroot(x+1)+2)
    lim x-->3 (x+1)-4 / (x-3)(squareroot(x+1)+2)
    lim x-->3 (x-3) / (x-3)(squareroot(x+1)+2) and I cancelled the (x-3) from top and bottom

    I then plugged 3 into (squareroot(x+1)+2) to get (squareroot(4)+2) which is 2+2 = 4

    However, it says the answer is wrong. I'm really not sure why.

    Help is appreciated! Thanks.
    \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt {x + 1}  - 2}}<br />
{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {\sqrt {x + 1}  - 2} \right)\left( {\sqrt {x + 1}  + 2} \right)}}<br />
{{\left( {x - 3} \right)\left( {\sqrt {x + 1}  + 2} \right)}} =

    = \mathop {\lim }\limits_{x \to 3} \frac{{x + 1 - 4}}<br />
{{\left( {x - 3} \right)\left( {\sqrt {x + 1}  + 2} \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}<br />
{{\left( {x - 3} \right)\left( {\sqrt {x + 1}  + 2} \right)}} =

    = \mathop {\lim }\limits_{x \to 3} \frac{1}{{\sqrt {x + 1}  + 2}} = \frac{1}{{\sqrt {3 + 1}  + 2}} = \frac{1}{4}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    32
    Oh wow, I feel silly! Thanks for showing me.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find the limit
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 24th 2011, 02:17 PM
  2. Replies: 12
    Last Post: August 26th 2010, 10:59 AM
  3. How to find this limit?
    Posted in the Calculus Forum
    Replies: 8
    Last Post: March 31st 2010, 03:11 PM
  4. How to find this limit?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 5th 2009, 02:07 PM
  5. Replies: 15
    Last Post: November 4th 2007, 07:21 PM

Search Tags


/mathhelpforum @mathhelpforum